623. Add One Row to Tree

1. Description

Given the root of a binary tree and two integers val and depth, add a row of nodes with value val at the given depth depth.
Note that the root node is at depth 1.
The adding rule is:

• Given the integer depth, for each not null tree node cur at the depth depth - 1, create two tree nodes with value val as cur’s left subtree root and right subtree root.
• cur’s original left subtree should be the left subtree of the new left subtree root.
• cur’s original right subtree should be the right subtree of the new right subtree root.
• If depth == 1 that means there is no depth depth - 1 at all, then create a tree node with value val as the new root of the whole original tree, and the original tree is the new root’s left subtree.

2. Example

Example 1:

Input: root = [4,2,6,3,1,5], val = 1, depth = 2
Output: [4,1,1,2,null,null,6,3,1,5]

Example 2:

Input: root = [4,2,null,3,1], val = 1, depth = 3
Output: [4,2,null,1,1,3,null,null,1]

3. Constraints

• The number of nodes in the tree is in the range [1, $10^4$].
• The depth of the tree is in the range [1, $10^4$].
• -100 <= Node.val <= 100
• $-10^5$ <= val <= $10^5$
• 1 <= depth <= the depth of tree + 1

4. Solutions

Breadth-First Search & Queue

n is the number of nodes in root Time complexity: O(n)
Space complexity: O(n)

class Solution {
public:
    TreeNode *addOneRow(TreeNode *root, int val, int depth) {
        if (depth == 1) {
            return new TreeNode(val, root, nullptr);
        }

        queue<TreeNode *> nodes{{root}};
        for (int i = 2; i < depth; ++i) {
            int level_size = nodes.size();
            for (; level_size > 0; --level_size) {
                auto node = nodes.front();
                nodes.pop();

                if (node->left != nullptr) {
                    nodes.push(node->left);
                }
                if (node->right != nullptr) {
                    nodes.push(node->right);
                }
            }
        }

        while (!nodes.empty()) {
            auto node = nodes.front();
            nodes.pop();

            auto new_left_node = new TreeNode(val, node->left, nullptr);
            node->left = new_left_node;

            auto new_right_node = new TreeNode(val, nullptr, node->right);
            node->right = new_right_node;
        }

        return root;
    }
};